Bit Pattern Problem

[fb]

Hi Everyone. Thanks for the help with the qudratic equation problem...I
didn't think about actually doing the math...whoops. Anyway... I'm
having some trouble getting the following program to work. I want to
output a bit pattern from base 10 input. All I get is a zero after the
input...I've looked over the code but can't see the problem...any ideas?
/* display the bit pattern corresponding to
a signed decimal integer */

#include<stdio. h>
#include<stdlib .h>

int main(void)
{
int a, b, m, count, nbits;
unsigned mask;

/* determine the word size in bits and set the initial mask */
nbits = 8 * sizeof(int);
m = 0x1 << (nbits - 1); /* Place 1 in leftmost position */

/* main loop */
do {
/* read a signed integer */
printf("\n\nEnt er an integer value (0 to stop): ", a);
scanf("%d", &a);

/* output the bit pattern */
mask = m;
for(count = 1; count <= nbits; count++); {
b = (a & mask) ? 1 : 0; /* set display bit on or off */
printf("%x", b); /* print display bit */
if (count % 4 == 0)
printf(" "); /* blank space after ever 4th digit */
mask >>= 1; /* shift mask 1 position to the right */
}
} while (a != 0);
return EXIT_SUCCESS; /* If it's there why not use it people! */
}

[Derrick Coetzee]

fb wrote:

nbits = 8 * sizeof(int);

The error is on this line.
--
Derrick Coetzee
I grant this newsgroup posting into the public domain. I disclaim all
express or implied warranty and all liability. I am not a professional.

[fb]

Derrick Coetzee wrote:

fb wrote:

nbits = 8 * sizeof(int);

The error is on this line.

right-o thanks.

[Derrick Coetzee]

Derrick Coetzee wrote:

fb wrote:

nbits = 8 * sizeof(int);

The error is on this line.

Oops, no it's not. I'm an idiot. The actual error is on this line:
for(count = 1; count <= nbits; count++); {

Watch out for those accidental semicolons! ^

--
Derrick Coetzee
I grant this newsgroup posting into the public domain. I disclaim all
express or implied warranty and all liability. I am not a professional.

[fb]

Derrick Coetzee wrote:

Derrick Coetzee wrote:

fb wrote:

nbits = 8 * sizeof(int);

The error is on this line.

Oops, no it's not. I'm an idiot. The actual error is on this line:

> for(count = 1; count <= nbits; count++); {

Watch out for those accidental semicolons! ^

cool...ok. ty.

[Mac]

On Wed, 01 Sep 2004 22:35:48 -0400, Derrick Coetzee wrote:

fb wrote:

nbits = 8 * sizeof(int);

The error is on this line.

Well, the line shouldn't be written that way, but I doubt that is the
problem. If anything, it should be:

nbits = CHAR_BIT * sizeof (int);

--Mac

[Mac]

On Thu, 02 Sep 2004 01:21:47 +0000, fb wrote:

Hi Everyone. Thanks for the help with the qudratic equation problem...I
didn't think about actually doing the math...whoops. Anyway... I'm
having some trouble getting the following program to work. I want to
output a bit pattern from base 10 input. All I get is a zero after the
input...I've looked over the code but can't see the problem...any ideas?
/* display the bit pattern corresponding to
a signed decimal integer */

#include<stdio. h>
#include<stdlib .h>

int main(void)
{
int a, b, m, count, nbits;
unsigned mask;

/* determine the word size in bits and set the initial mask */
nbits = 8 * sizeof(int);
m = 0x1 << (nbits - 1); /* Place 1 in leftmost position */

/* main loop */
do {
/* read a signed integer */
printf("\n\nEnt er an integer value (0 to stop): ", a);
scanf("%d", &a);

/* output the bit pattern */
mask = m;
for(count = 1; count <= nbits; count++); {
Someone else already pointed out that you didn't mean to put a semicolon
after your for loop. That collapses it to an empty loop.
b = (a & mask) ? 1 : 0; /* set display bit on or off */
printf("%x", b); /* print display bit */
if (count % 4 == 0)
printf(" "); /* blank space after ever 4th digit */
mask >>= 1; /* shift mask 1 position to the right */
I don't think the above line will do what you expect.

From the last public draft of c89:

3.3.7 Bitwise shift operators
[...]
The result of E1 >> E2 is E1 right-shifted E2 bit positions.
[...]
If E1 has a signed type and a negative
value, the resulting value is implementation-defined.

At the machine level, you sometimes see a shift with sign extension for
signed integers, and your implementation might use that type of shift
rather than a logic shift. What this means is that shifting a negative
number to the right may fill the higher order bits with 1's. The
documentation for your implementation should say what it does, though.

As far as I know, which may not be very far, every implementation will
probably either fill the high order bits with zeros or fill them with ones.
} } while (a != 0);
return EXIT_SUCCESS; /* If it's there why not use it people! */
}

--Mac

[CBFalconer]

fb wrote:

.... snip ... I'm having some trouble getting the following program to work.
I want to output a bit pattern from base 10 input. All I get is
a zero after the input...I've looked over the code but can't see
the problem...any ideas?

/* display the bit pattern corresponding to
a signed decimal integer */

#include<stdio. h>
#include<stdlib .h>

int main(void)
{
int a, b, m, count, nbits;
unsigned mask;

/* determine the word size in bits and set the initial mask */
nbits = 8 * sizeof(int);
You should #include <limits.h> and replace 8 by CHAR_BIT. Bytes
need not contain 8 bits.
m = 0x1 << (nbits - 1); /* Place 1 in leftmost position */
To do this m should be an unsigned int. Shifting into the sign
position results in undefined behavior. Unsigned ints don't have
a sign bit. Make all the variables unsigned.

/* main loop */
do {
/* read a signed integer */
printf("\n\nEnt er an integer value (0 to stop): ", a); ^^^
What is this? You also need a "fflush(stdin); " after this.
scanf("%d", &a);
Check the result from scanf, which should be 1. Maybe:
if (1 != scanf("%ud", &a)) break;

/* output the bit pattern */
mask = m;
for(count = 1; count <= nbits; count++); { ^
What's this semicolon for?
b = (a & mask) ? 1 : 0; /* set display bit on or off */
printf("%x", b); /* print display bit */
simpler: putc(b + '0', stdout);
if (count % 4 == 0)
printf(" "); /* blank space after ever 4th digit */
simpler: putc(' ', stdout);
mask >>= 1; /* shift mask 1 position to the right */
}
} while (a != 0);
return EXIT_SUCCESS; /* If it's there why not use it people! */
}

--
A: Because it fouls the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

[Gordon Burditt]

>> /* read a signed integer */

printf("\n\nEnt er an integer value (0 to stop): ", a);

^^^
What is this? You also need a "fflush(stdin); " after this.

No, you absolutely NEVER need a "fflush(stdin)" . If you absolutely
have to invoke undefined behavior, use fflush(void main()). It will
probably save you a lot of time by failing to compile.

Gordon L. Burditt

[Mike Wahler]

"CBFalconer " <cb********@yah oo.com> wrote in message
news:41******** *******@yahoo.c om...

/* main loop */
do {
/* read a signed integer */
printf("\n\nEnt er an integer value (0 to stop): ", a);

^^^
What is this? You also need a "fflush(stdin); " after this.

Make that:

fflush(stdout);

But you knew that. :-)

-Mike